Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set - Page 646: 24

Answer

$a_{1} = 75, r= \frac{1}{5}$

Work Step by Step

Given $a_{2} = 15$ $a_{3} = 3$ In geometric progression, common ratio, $r = \frac{a_{n}}{a_{n-1}}$ $r = \frac{a_{3}}{a_{2}} = \frac{3}{15} = \frac{1}{5}$ $a_{2} = a_{1} . r^{2-1} $ using $a_{n} = a_{1} . r^{n-1} $ $a_{2} = a_{1} . r^{1} $ Using $r$ and $a_{2}$ values, $15 = a_{1} .(\frac{1}{5})^{1} $ $15 = a_{1} .(\frac{1}{5}) $ $a_{1} = 15 \times 5 $ $a_{1} = 75$
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